Calculation 1

Example 1

For Example 2 See: Calculation 2
This example is used to explain the mathematics involved in length contraction

Consider a spaceship with a length l0 = 300000 km
This spaceship has a speed v = 0.999*c
There is one observer at the back of the spaceship and one at the front of the space ship.
At the back of the spaceship a light pulse is transmitted to the front and reflected to the back.

Questions:

In the rest frame, when will the pulse reach the front of the spaceship?
In the rest frame, when will the pulse reach the back of the spaceship?
In the moving frame, when will the pulse reach the front of the spaceship?
In the moving frame, when will the pulse reach the back of the spaceship?
With v=0, when will the pulse reach the front of the spaceship?
With v=0, when will the pulse reach the back of the spaceship?
What if no length contraction is involved?

In questions 1 and 2 clocks in the rest frame are used. In questions 3 and 4 clocks in the moving spaceship are used.

     |               |                          /           /
t1+t2.               |                    t1+t2.           /
     | .             |                        /  .        /
     |   .           |                       /     .     /
     |     .         |                      /        .  /
     |       .       |                     /           .t1
     |         .     |                    /          ./
     |           .   |                   /         . /
     |             . |                  /        .  /
     |               .t1               /       .   /
     |             . |                /      .    /
     |           .   |               /     .     /
     |         .     |              /    .      /
     |       .       |             /   .       .t4
     |     .         |            /  .       ./
     |   .           |           / .       . /
     | .             |          /.       .  /   ---->
---t3.---------------.t4-----t3.-------.---/---t0
     | .           . |        /  .   .    /
     |   .       .   |    ---B-----.-----F---
     |     .   .     |
-----B-------.-------F-           
            
           v=0                  v>0
         
           B = Back of the train or spaceship
           F = Front of the train or spaceship
           . = Light signal

The above left figure shows the situation for v = 0 (The questions 5 and 6)
The above right figure shows two aspects of this example for v>0 (The questions 1-4)
Each figure is subdivided into two parts.

The top part above the line t0 shows the example:
- At t0 (=t3) there is a light signal at the back of the train.
- This light signal reaches the front F at the train at t1.
- The light signal reaches the back B at the train at t1+t2
- Delta time between the back and the front signals is t2
The bottom part shows clock synchronization.
- Clock synchronization starts with two light signals at the middle from the train.
- One signal reaches the back B of the train at t3. This signal is the start signal t0 of the example.
- The other signal reaches the front B of the train at t4.
- The two signals t3 and t4 are used by the observers on the train to reset their watches.

Answers

First of all length contraction is involved.
That means the length of the moving train is not l0 but l (measured in the rest frame).
l = l0 * sqrt(1-v*v/c*c) = l0 / gamma
gamma = 1 / sqrt(1-v*v/c*c)
The last sixth question is the easiest:
Total duration in rest frame with speed is zero = 2 * l0 / c = 2 Secs.
The answer for the fifth question is:
The duration going from back to front is 1 Second.
In order to answer the first question you must use the formula:
l + v*t = c*t or
l = c*t - v*t or
t1 = t = l / (c-v)
In order to answer the second question you must use the formula:
l = v*t + c*t or
t2 = t = l / (c+v)
The total time since t0 = 0 is equal to t1+t2
t1+t2 = l / (c-v) + l / (c+v)
In order to answer the fourth question you must divide the previous time t1+t2 by gamma. This takes Time Dilation into account i.e. that the moving clock runs slower.
In order to answer the third question you must first synchronise the two clocks on board of the train.
The time t3 that one signal reaches the front = 0.5*l/(c+v)
The time t4 that one signal reaches the back = 0.5*l/(c-v)
If the watch of the observer at F is reseted at t4 then the time from the back of the train does not reach the observer at t1 but at t1-(t4-t3) in rest frame
In the reference frame of the moving train you have to divide that value by gamma

In order to do the example for a particular value of v/c you can use the following calculator:

v/c	v	c	gamma

l0	l=l0/gamma

t1	t2	t1+t2	/gamma

t3	t4	t4-t3	t1-(t4-t3)	/gamma

Calculations when NO length contraction is involved.

l=l0 t1 t2 t1+t2 /gamma

The first line shows the value v/c in red box.
This value can be modified:
Enter a value and select any other point on this diplay in order to excute the calculations.

Enter 0.8, 0.99, 0.999 and 0.

The other values are resp: the speed v, the speed of light c and gamma
The second line shows the rest length l0 of the train and the length of the moving train l
The third line shows the answers on the questions 1, 2 and 4.
The answer on question 1 is t1, on 2 is t1+t2 and on 4 is the value (t1+t2)/gamma
Specific the answer on question 4 is important because it means that the length of the rod, at rest in the rest frame, measured in the rest frame (question 5), is identical as the length of the moving rod, measured in the moving frame (question 4).
The fourth line shows the answer on the question 3.
The fifth line shows the answer on the questions 1, 2 and 4 when NO length contraction is involved.
The answer on question 1 is t1, on 2 is t1+t2 and on 4 is the value (t1+t2)/gamma

Experimental Verification of Calculations

The idea behind this example is to perform five "identical" experiments and to predict their outcome.

In order to measure the length of a rod at rest an Observer sends a light signal to front of the rod. At the front of the rod there is a mirror. The reflection time is 2 seconds that means the length l0 of the rod is c light seconds. The reflection time is the time t1+t2 in the picture above, at the left.
Next we study a moving rod with speed v and measure its length. We do that by sending a light signal to the front of the rod when the back end is with the Observer. The reflection time is measured by the Observer at rest. The reflection time is the value 2*t1 in the picture at the right, above. There are two possibilities.
- Without Length Contraction: 2*l0 /(c-v)
- With Length Contraction: 2*l0 / (c-v) / gamma = 2 * l / (c-v)
Which is correct?
Next we repeat the same experiment with the only exception that the reflection time is measured by a moving Observer at the back end of the rod. The reflection time is measured with clocks (synchronized) in the rest frame. The reflection time is the value t1+t2 in the picture at the right, above. There are again two possibilities:
- Without Length Contraction: l0 /(c-v) + l0 /(c+v)
- With Length Contraction: {l0 /(c-v) + l0 /(c+v)} / gamma = l /(c-v) + l /(c+v)
Which is correct?
Next we repeat the same experiment with the only exception that the reflection time is measured by a moving Observer at the back end of the rod. The reflection time is measured with a moving clock. The reflection time is the value t1+t2 in the picture at the right, above. There are now four possibilities:
- Without Length Contraction and without Time Dilation: l0 /(c-v) + l0 /(c+v)
- With Length Contraction and without Time Dilation: {l0 /(c-v) + l0 /(c+v)} / gamma = l /(c-v) + l /(c+v)
- Without Length Contraction and with Time Dilation: {l0 /(c-v) + l0 /(c+v)} / gamma
- With Length Contraction and with Time Dilation: [{l0 /(c-v) + l0 /(c+v)} / gamma ] / gamma =
  {l0 /(c-v) + l0 /(c+v)} * (1-v*v/c*c) = {l0*(c+v) + l0*(c-v)}/{(c-v)*(c+v)} * {(c*c-v*v}/c*c) =
  {l0*(c+v) + l0*(c-v)}/c*c) = 2 * l0 * c / c² = 2 * l0 / c = 2 seconds
Which is correct?

Accordingly to SR: it is the last possiblity.
However there is one problem: you can not perform the above experiments which enough accuracy to claim that Special Relativity is true.

The Laws of Physics are the same in all inertial frames

The laws of physics look the same in all inertial reference frames.

See: http://www.pa.uky.edu/~cvj/as500_lec3/as500_lec3.html

Consider the next questions and issues:

Consider a mirror at distance l and an observer O1 at rest with a clock.
Reflection time of single photon t1 = 2*l/c.
Consider a moving observer O2 towards mirror at speed v.
Observer O2 will meet the reflected photon at t2 in rest frame at point X.
Distance traveled by observer is l2 = v * t2
Distance traveled by photon is l + (l-l2) = 2*l - l2
t2 = l2/v = (2*l - l2) /c
l2 = 2 * v * l /(c+v)
t2 = l2 / v = 2 * l /(c+v)
```
        l2               l-l2         |
 O1----->----X------------><----------| Mirror  
   -----O2--->                        |
```
Consider the moving observer also has a moving clock. In order to calculate the time of the moving clock you have to divide t2 by gamma. This takes Time dilation into account
t2* = 2 * l/(c+v)/gamma = 2*l * sqrt (1-v*v/c*c)/(c+v) =
2*l/c * sqrt (c*c-v*v)/(c+v)² = 2*l/c * sqrt (c-v)/(c+v)
t2* is not equal to t2.
Implying that the results of the same test for the observer at rest (time t2) is not the same for the moving observer (time t2*)
This seems to be in conflict with the basic principles of Special Relativity.

v/c	v	c	gamma

l	t1	t2	t2*

Reflection

The last value of line three and the last value of line four demonstrates, that the observations of the observers in the moving train are identical for those observers as when the train stands still.
The observations are independent of the speed v of the train, in accordance with Special Relativity.

Line five, the values when no Length Contraction is involved, is not in accordance with Special Relativity.

Physical Interpretation

If the results of actual experiments are in accordance with the lines 1 to 4 than this implies that Length Contraction is a physical effect.
The same for Time Dilation.
If the results of experiments and or observations are in accordance with the line 5 than no Length Contraction is involved.

Feedback

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Created: 8 February 2002
Modified: 19 November 2004

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